A337737 Least number k such that there are exactly n cubefull numbers between k^3 and (k+1)^3.

1, 2, 6, 15, 12, 25, 43, 73, 480, 1981, 3205, 9038, 16099, 376340, 211318, 2461230, 2253517, 16907618, 106308537, 312911063

Offset: 0

Author

Amiram Eldar, Sep 17 2020

Keywords

• nonn
• more

Comments

a(n) = least k such that A337736(k) = n.

Shiu (1991) proved that infinitely many values of k exist for every n. Therefore, this sequence is infinite.

Examples

			a(0) = 1 since there are no cubefull numbers between 1^3 = 1 and 2^3 = 8.
a(1) = 2 since there is one cubefull number, 16 = 2^4, between 2^3 = 8 and 3^3 = 27.
a(2) = 6 since there are 2 cubefull numbers, 243 = 3^5 and 256 = 2^8, between 6^3 = 216 and 7^3 = 343.
		

Links

• P. Shiu, The distribution of cube-full numbers, Glasgow Mathematical Journal, Vol. 33, No. 3 (1991), pp. 287-295. See section 3, p. 291.

Crossrefs

Cf. A000578, A036966, A119242, A337736.

Programs

Mathematica

cubQ[n_] := Min[FactorInteger[n][[;; , 2]]] > 2; f[n_] := Count[Range[n^3 + 1, (n + 1)^3 - 1], _?cubQ]; mx = 8; s = Table[0,{mx}]; c = 0; n = 1; While[c < mx, i = f[n] + 1; If[i <= mx && s[[i]] == 0, c++; s[[i]] = n]; n++] ;s

Python

from math import gcd
from sympy import integer_nthroot, factorint
def A337737(n):
    if n == 0: return 1
    a, k = 0, 1
    while True:
        m, c = k**3, 0
        for x in range(1,integer_nthroot(m,5)[0]+1):
            if all(d<=1 for d in factorint(x).values()):
                for y in range(1,integer_nthroot(z:=m//x**5,4)[0]+1):
                    if gcd(x,y)==1 and all(d<=1 for d in factorint(y).values()):
                        c += integer_nthroot(z//y**4,3)[0]
        if c-a-1 == n:
            return k-1
        k += 1
        a = c # Chai Wah Wu, Apr 23 2025

Extensions

a(12)-a(16) from David A. Corneth, Sep 18 2020

a(17)-a(19) from Bert Dobbelaere, Sep 19 2020